Question

Ksp for \( \text{Cr(OH)}_3 \) is \( 1.6 \times 10^{-30} \). What is the molar solubility of this salt in water?

• A. \( s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}} \)
• B. \( \frac{1.8 \times 10^{-30}}{27} \)
• C. \( \sqrt[5]{1.8 \times 10^{-30}} \)
• D. \( \frac{2 \sqrt{1.6 \times 10^{-30}}}{27} \)

Hint:

When solving for solubility from \( K_{{sp}} \), set up the equation for \( K_{{sp}} \) in terms of the molar solubility \( s \), and solve for \( s \) using the appropriate algebraic steps.

Answer 1

The Correct Option is A

To find the molar solubility of \( \text{Cr(OH)}_3 \) in water, we begin by understanding the dissolution equation and the relationship between solubility product (\( K_{sp} \)) and molar solubility (\( s \)). 

The dissolution equilibrium of \( \text{Cr(OH)}_3 \) in water can be given by:

\(\text{Cr(OH)}_3 (s) \rightleftharpoons \text{Cr}^{3+} (aq) + 3\text{OH}^- (aq)\)

The expression for the solubility product \( K_{sp} \) is:

\(K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3\)

If \( s \) is the molar solubility of \( \text{Cr(OH)}_3 \), then the concentration of \( \text{Cr}^{3+} \) will be \( s \) and the concentration of \( \text{OH}^- \) will be \( 3s \). Substituting these into the expression for \( K_{sp} \):

\(K_{sp} = (s)(3s)^3 = 27s^4\)

Given \( K_{sp} = 1.6 \times 10^{-30} \), we substitute this value into the equation:

\(1.6 \times 10^{-30} = 27s^4\)

Solving for \( s \), we get:

\(s^4 = \frac{1.6 \times 10^{-30}}{27}\)

Taking the fourth root of both sides gives:

\(s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)

Thus, the molar solubility of \( \text{Cr(OH)}_3 \) in water is given by the expression:

\(s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)

Answer 2

The dissolution of \( {Cr(OH)}_3 \) is represented as: \[ {Cr(OH)}_3 (s) \rightleftharpoons {Cr}^{3+} (aq) + 3 \, {OH}^- (aq) \] Let the molar solubility of \( {Cr(OH)}_3 \) be \( s \). Therefore, the concentration of \( {Cr}^{3+} \) is \( s \) and the concentration of \( {OH}^- \) is \( 3s \). 
The solubility product expression is: \[ K_{{sp}} = [{Cr}^{3+}][{OH}^-]^3 = s(3s)^3 = 27s^4 \] Given \( K_{{sp}} = 1.6 \times 10^{-30} \), we can solve for \( s \): \[ 1.6 \times 10^{-30} = 27s^4 \] \[ s^4 = \frac{1.6 \times 10^{-30}}{27} \] \[ s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}} \] 
Thus, the molar solubility is \[ s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}} \].